Let $R$ and $S$ be two commutative rings with unity. Prove that $R\times S$ is NOT an integral domain.

This is the best I could think of so far, please give me a push in the right direction and correct me.

It suffices to show that $R\times S$ has zero-divisors

Therefore, let $a$ be an element of $R$ and $b$ be an element of $S$, thus $R\times S$ has elements of the form $(a,b)$.

Consider the elements $A=(a,0)$ and $B=(0,b)$, such that $a$ does not equal zero and $b$ does not equal zero. Then $AB = 0$. Therefore $R\times S$ has zero divisors.

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    $\begingroup$ Your argument is correct, but what is your question? $\endgroup$ – Crostul May 11 '15 at 18:06
  • $\begingroup$ the question was proving that R+S is not an integral domain? I wanted to know if my proof and logic was correct $\endgroup$ – B ry May 11 '15 at 18:08
  • $\begingroup$ Your argument is correct then. $\endgroup$ – jgon May 11 '15 at 18:09
  • $\begingroup$ @Bry If you bothered to use the search function, you would have found confirmation at a question entitled "Prove R×R is NOT an integral domain." $\endgroup$ – rschwieb May 11 '15 at 19:04
  • $\begingroup$ I used it and I guess I didn't realize RxR is the same as external direct product $\endgroup$ – B ry May 11 '15 at 19:25

Your argument here is perfectly correct, as others have indicated.

It would help the clarity of your proof (and, on a fairly pedantic level, would make your proof more correct) if you chose specific non-zero elements $a,b$. In particular, we could take $a = 1_R$ and $b = 1_S$.

This way, we could always say that we're explicitly using the fact that a ring with unity has both a $1$ and a $0$.

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    $\begingroup$ Indeed, $1_R$ and $1_S$ are the only elements which we know to exist and be $\ne 0$ in the given rings $\endgroup$ – Hagen von Eitzen May 11 '15 at 18:13
  • $\begingroup$ Awesome, thank you for helping me! $\endgroup$ – B ry May 11 '15 at 18:14
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    $\begingroup$ Sorry to be pedantic, but I don't think $1 \not= 0$ is usually taken as one of the ring axioms. For once, Wikipedia bears me out: en.wikipedia.org/wiki/… en.wikipedia.org/wiki/Zero_ring $\endgroup$ – Rob Arthan May 11 '15 at 21:16

yes your answer is indeed correct take two elements that are not zero than show that their product is zero.


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